Question

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of
elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, then
find the height of the building,​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Height\:of\:building=20\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high.

• We have to find the height of the building.

$\underline \bold{Given : } \\ \implies angle \: of \: elevation \: of \: top \: of \: \\ \: \: \: \: \: \: \: \: \: \: \: building \: from \: foot \: of \:tower \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \theta_{1} = 30 \degree \\ \\ \implies angle \: of \: elevation \: of \: top \: of \: \\ \: \: \: \: \: \: \: \: \: \: \: tower \: from \: foot \: of \: building \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \theta_{2} = 60 \degree \\ \\ \implies height \: of \: tower = 60 \: m \\ \\ \underline \bold{To \: Find : } \\ \implies height \: of \: building = ?$

• According to given question :

$\bold{For \: angle \: of \: elevation = \theta_{1}} \\ \implies tan \theta_{1} = \frac{p}{b} \\ \\ \implies tan \: 30 \degree = \frac{Height \: of \: building}{Distance \: between \: tower \: and \: building} \\ \\ \implies \frac{1}{\sqrt{3} } = \frac{x}{l} \\ \\ \implies l = \sqrt{3}x - - - - - (1) \\ \\ \bold{For \: angle \: of \: elevation = \theta_{2}} \\ \implies tan \theta_{2} = \frac{Height \: of \: tower}{Distance \: between \: tower \: and \: building} \\ \\ \implies \sqrt{3} = \frac{60}{l} \\ \\ \bold{putting \: value \: of \: l \: from \: (1)} \\ \implies \sqrt{3} = \frac{60}{ \sqrt{3}x } \\ \\ \implies \sqrt{3} \times \sqrt{3} x = 60 \\ \\ \implies x = \frac{60}{3} \\ \\ \bold{ \implies x = 20 \: m} \\ \\ \bold{\therefore Height \: of \: building = 20 \: m}$